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Difficulty:
95%
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Question Stats:
51% (02:02) correct
49%
(02:36)
wrong
based on 291
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If x is a non-negative number and \(x ≠ 3\sqrt{x} + 4\), which of the following must be true?
I. (x - 1)(x - 16) ≠ 0
II. |17 - x| ≠ 1
III. x is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
M46-10
I. (x - 1)(x - 16) ≠ 0
II. |17 - x| ≠ 1
III. x is not a square of an integer
A. I only
B. II only
C. III only
D. I, and II only
E. None of the above
M46-10
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18 Dec 2025, 10:00
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Bunuel
Let's solve \(x = 3\sqrt{x} + 4\).
Let \(\sqrt{x}= t\), so \(x = t^2\).
Substitute:
\(t^2 = 3t + 4\)
\(t^2 - 3t - 4 = 0\)
Factor:
\((t - 4)(t + 1) = 0\)
So \(t = 4\) or \(t = -1\).
But \(t = \sqrt{x}\) cannot be negative, so only \(t = 4\) is valid.
Thus \(\sqrt{x} = 4\), so \(x = 16\).
Therefore \(x ≠ 3\sqrt{x} + 4\) means \(x ≠ 16\). That is the only restriction (along with x being a non-negative number).
Now test each statement to see whether it must be true when \(x ≠ 16\).
I. (x - 1)(x - 16) ≠ 0
This expression is nonzero unless \(x = 1\) or \(x = 16\). We only know \(x ≠ 16\), but x could be 1, which makes the expression 0. So Statement I is not guaranteed. Not always true.
II. |17 - x| ≠ 1
\(|17 - x| = 1\) corresponds to \(x = 16\) or \(x = 18\). We only know \(x ≠ 16\), but x could be 18, which would make \(|17 - x| = 1\). So Statement II is not guaranteed. Not always true.
III. x is not a square of an integer
We only know \(x ≠ 16\), but x could be 0, 1, 4, 9, 25, 36, etc. So Statement III is not guaranteed. Not always true.
Answer: E.
General Discussion
18 Dec 2025, 09:07
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x is non-negative and x not equals 3 sqrt(x) + 4
Check when equality would happen:
Let sqrt(x) = t >= 0
so x = t^2
then t^2 = 3t +4
t^2-3t-4 = 0
(t-4)(t+1) = 0
since t>= 0
t = 4, so x =16
I. x not equals 1 and 16
but x = 1 is allowed.
II. x =16 and x =18
But x =18 is allowed
III. x is not a square of an integer
but x =9 is allowed
None of the statements must be true
Answer E
Check when equality would happen:
Let sqrt(x) = t >= 0
so x = t^2
then t^2 = 3t +4
t^2-3t-4 = 0
(t-4)(t+1) = 0
since t>= 0
t = 4, so x =16
I. x not equals 1 and 16
but x = 1 is allowed.
II. x =16 and x =18
But x =18 is allowed
III. x is not a square of an integer
but x =9 is allowed
None of the statements must be true
Answer E
